Math 2250-1
Mon Nov 12
10.5, EP7.6
Today we focus on the unit impulse ("delta function") Laplace transform entry, and the periodic function
table entry, from those shown below. On Friday we discussed using the unit step function to turn
functions on and off. We also briefly discussed convolution integrals, to find inverse Laplace transforms
of products F s G s . And, at the end of today's notes we'll see (EP7.6) why that convolution formula is
so important in applications.
f t % CeM t
f t with
N
for s O M
f t eKs t dt
Fs d
comments
0
u tKa
eKa s
s
unit step function
for turning components on and
off at t = a .
f tKa u tKa
eKa sF s
d tKa
eKa s
t
f t g t K t dt
more complicated on/off
unit impulse/delta "function"
convolution integrals to invert
Laplace transform products
Fs Gs
0
p
1
f t
with period p
1 K eKps
f t eKs t dt
for periodic forcing which is not
sinusoidal.
0
Periodic functions: Let f t be periodic with period p, i.e. f t C p = f t c t.
Notice that we can decompose f into the sum of a function which equals f on the interval 0, p and is zero
on p,N , with a function that is zero on 0, p and equals f on p,N :
f t = f t 1Ku tKp
= f t 1Ku tKp
Cu tKp f t
Cu tKp f tKp
(since f is pKperiodic).
If we take the Laplace transform of this identity use the definition of Laplace transform for the first term on
the right and the translation theorem for the second term, we get
p
F s =
f t eKs t dt C eKp s F s
0
C0 F s
1 K eKp s =
p
f t eKs t dt
0
F s =
p
1
f t eKs t dt .
Kp s
1Ke
0
Exercise 1) Find the Laplace transform of the function whose graph is shown below. The function has
period 4 :
1
0
0
(solution: F s =
2
1 K eKs
s 1 K eK4 s
4
)
t
6
8
10
EP 7.6 impulse functions and the d operator.
Consider a force f t acting on an object for only on a very short time interval a % t % a C e, for
example as when a bat hits a ball. This impulse p of the force is defined to be the integral
aCe
pd
f t dt
a
and it measures the net change in momentum of the object since by Newton's second law
m v# t = f t
aCe
aCe
m v# t dt =
0
a
f t dt = p
a
aCe
0mv t
t=a
=p.
Since the impulse p only depends on the integral of f t , and since the exact form of f is unlikely to be
known in any case, the easiest model is to replace f with a constant force having the same total impulse, i.e.
to set
f = p da, e t
where da, e t is the unit impulse function given by
da, e t =
0, t ! a
1
, a % t ! aCe
e
.
0,
t R aCe
Notice that
aCe
aCe
da, e t dt =
a
a
1
dt = 1 .
e
Here's a graph of d2, .1 t , for example:
10
8
6
4
2
0
0
1
2
t
3
4
Since the unit impulse function is a linear combination of unit step functions, we could solve differential
equations with impulse functions so-constructed. As far as Laplace transform goes, it's even easier to take
the limit as e/0 for the Laplace transformsL da, e t s , and this effectively models impulses on very
short time scales.
da, e t =
1
u tKa Ku tK aCe
e
0 L da, e t
eKa s
eK a C e
K
s
s
1
s =
e
s
1KeKe s
=e
.
es
In Laplace land we can use L'Hopital's rule (in the variable e) to take the limit as e/0:
Ke s
s eKe s
Ka s 1Ke
Ka s
lim e
=e
lim
= eKa s .
e/0
e/0
s
es
Ka s
The result in time t space is not really a function but we call it the "delta function" d t K a anyways, and
visualize it as a function that is zero everywhere except at t = a, and that it is infinite at t = a in such a way
that its integral over any open interval containing a equals one. As explained in EP7.6, the delta "function"
can be thought of in a rigorous way as an operator, not as a function. It can also be thought of as the
derivative of the unit step function u t K a , and this is consistent with the Laplace table entries for
derivatives of functions. In any case, this leads to the very useful Laplace transform table entry
d tKa
unit impulse function
eKa s
for impulse forcing
Exercise 2) Revisit the swing from Friday and solve the IVP below for x t . In this case the parent is
providing an impulse each time the child passes through equilibrium position after completing a cycle.
x## t C x t = .2 p d t C d t K 2 p C d t K 4 p C d t K 6 p C d t K 8 p
x 0 =0
x# 0 = 0 .
Impulse forcing of mechanical systems
EP 7.6
> with plots :
> plot1 d plot .1$t$sin t , t = 0 ..10$Pi, color = black :
plot2 d plot Pi$sin t , t = 10$Pi ..20$Pi, color = black :
plot3 d plot Pi, t = 10$Pi ..20$Pi, color = black, linestyle = 2 :
plot4 d plot KPi, t = 10$Pi ..20$Pi, color = black, linestyle = 2 :
plot5 d plot .1$t, t = 0 ..10$Pi, color = black, linestyle = 2 :
plot6 d plot K.1$t, t = 0 ..10$Pi, color = black, linestyle = 2 :
display plot1, plot2, plot3, plot4, plot5, plot6 , title = `Friday adventures at the swingset` ;
Friday adventures at the swingset
3
2
1
0
K1
K2
K3
2p
4p
6p
8p
10 p
t
12 p
14 p
16 p
18 p
20 p
>
impulse solution: five equal impulses to get same final amplitude of p meters - Exercise 2:
> f d t/.2$Pi$sum Heaviside t K k$2$Pi $sin t K k$2$Pi , k = 0 ..4 :
> plot f t , t = 0 ..20$Pi, color = black, title = `lazy parent on Monday` ;
lazy parent on Monday
3
2
1
0
K1
2p
4p
6p
8p
K2
K3
10 p
t
12 p
>
Or, an impulse at t = 0 and another one at t = 10 p .
> g d t/.2$Pi$ 2$sin t C 3$Heaviside t K 10$Pi $sin t K 10$Pi
> plot g t , t = 0 ..20$Pi, color = black, title = `very lazy parent` ;
14 p
16 p
18 p
20 p
14 p
16 p
18 p
20 p
:
very lazy parent
3
2
1
0
K1
K2
K3
>
2p
4p
6p
8p
10 p
t
12 p
Convolutions and solutions to non-homogeneous physical oscillation problems (EP7.6 p. 499-501)
Consider a mechanical or electrical forced oscillation problem for x t , and the particular solution that
begins at rest:
a x##C b x#C c x = f t
x 0 =0
x# 0 = 0 .
Then in Laplace land, this equation is equivalent to
a s2 X s C b s X s C c X s = F s
0 X s a s2 C b s C c = F s
1
0X s =F s $ 2
dF s W s .
a s Cb sCc
LK1 W s
t = w t is called the weight function for the physical system, and because of the
convolution table entry
t
f t g t K t dt
Fs Gs
convolution integrals to invert
Laplace transform products
0
the solution is given by
t
x t = f)w t = w)f t =
w t f t K t dt .
0
This idea generalizes to much more complicated mechanical and circuit systems, and is how engineers
experiment mathematically with how proposed configurations will respond to various input forcing
functions, once they figure out the weight function for their system.
Exercise 3. Let's play the resonance game, with non-sinusoidal forcing functions. We'll stick with our
earlier swing, but consider various forcing periodic functions f t that we haven't thought about before.
x## t C x t = f t
x 0 =0
x# 0 = 0
a) Find the weight function w t .
b) Write down the solution formula for x t as a convolution integral.
c) Play the resonance game on the following pages ...
We worked out that the solution to our DE IVP will be
t
x t =
sin t f t K t dt
0
Since the unforced system has a natural angular frequency w0 = 1 , we expect resonance when the forcing
2p
= 2 p. We will discover that there is the possibility
w0
for resonance if the period of f is a multiple of T0 . (Also, forcing at the natural period doesn't guarantee
resonance...it depends what function you force with.)
function has the corresponding period of T0 =
Example 1) A square wave forcing function with amplitude 1 and period 2 p. Let's talk about how we
came up with the formula (which works until t = 11 p ) .
> with plots :
10
> f1 d t/K1 C 2$
>
K1 n $Heaviside t K n$Pi
:
n=0
plot1a d plot f1 t , t = 0 ..30, color = green :
display plot1a, title = `square wave forcing at natural period` ;
square wave forcing at natural period
1
0
10
K1
20
30
t
1) What's your vote? Is this square wave going to induce resonance, i.e. a response with linearly growing
amplitude?
t
> x1 d t/ sin t $f1 t K t dt :
0
plot1b d plot x1 t , t = 0 ..30, color = black :
display plot1a, plot1b , title = `resonance response ?` ;
Example 2) A triangle wave forcing function, same period
t
> f2 d t/ f1 s ds K 1.5 : # this antiderivative of square wave should be triangle wave
0
plot2a d plot f2 t , t = 0 ..30, color = green :
display plot2a, title = `triangle wave forcing at natural period` ;
triangle wave forcing at natural period
1
0
K1
10
20
30
t
2) Resonance?
t
> x2 d t/ sin t $f2 t K t dt :
0
plot2b d plot x2 t , t = 0 ..30, color = black :
display plot2a, plot2b , title = `resonance response ?` ;
Example 3) Forcing not at the natural period, e.g. with a square wave having period T = 2 .
20
> f3 d t/K1 C 2$
>
K1 n $Heaviside t K n :
n=0
plot3a d plot f3 t , t = 0 ..20, color = green :
display plot3a, title = `out of phase square wave forcing` ;
out of phase square wave forcing
1
0
K1
5
10
t
3) Resonance?
t
> x3 d t/ sin t $f3 t K t dt :
0
plot3b d plot x3 t , t = 0 ..20, color = black :
display plot3a, plot3b , title = `resonance response ?` ;
15
20
Example 4) Forcing not at the natural period, e.g. with a particular wave having period T = 6 p .
10
> f4 d t/
>
Heaviside t K 6$ n$p K Heaviside t K 6$n C 1 $p
:
n=0
plot4a d plot f4 t , t = 0 ..150, color = green :
display plot4a, title = sporadic square wave with period 6p ;
sporadic square wave with period 6p
1
0
0
50
t
4) Resonance?
t
> x4 d t/ sin t $f4 t K t dt :
0
plot4b d plot x4 t , t = 0 ..150, color = black :
display plot4a, plot4b , title = `resonance response ?` ;
>
100
150
Hey, what happened???? How do we need to modify our thinking if we force a system with something
which is not sinusoidal, in terms of worrying about resonance? In the case that this was modeling a swing
(pendulum), how is it getting pushed?
Precise Answer: It turns out that any periodic function with period P is a (possibly infinite) superposition
P P P
of a constant function with cosine and sine functions of periods P,
,
,
,... . Equivalently, these
2 3 4
functions in the superposition are
2$p
1, cos w t , sin w t , cos 2 w t , sin 2 w t , cos 3 w t , sin 3 w t ,... with ω =
. This is the
P
theory of Fourier series, which you will study in other courses, e.g. Math 3150, Partial Differential
Equations. If the given periodic forcing function f t has non-zero terms in this superposition for which
P
2$p
n$w = w0 (the natural angular frequency) (equivalently
=
), there will be resonance; otherwise, no
n
w0
resonance. We could already have understood some of this in Chapter 5, for example
Exercise 4) The natural period of the following DE is (still) T0 = 2 p . Which of the two DE's below will
have solutions that exhibit resonance? Use Chapter 5 superposition ideas. Compare your answer with
how the period of the forcing function compares to the natural period of the system.
a)
t
x## t C x t = cos t C sin
.
3
b)
x## t C x t = cos 2 t K 3 sin 3 t .